Other Single-Loop Circuits — Question 5

Section 27-4 · Physics pages 4380–4381

Question 5 A monotone nonlinear load gives one physical operating current An ideal 14.0 V source drives a 1.00 ohm resistor and a nonlinear load with Vd(I) = (2.00 Ω)I + (1.00 V/A3)I3. + − 14 V R nonlinear I Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find the steady current and prove uniqueness. (b) Find the small current change for a small emf change deltaE near that state.
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Question 5 - Solution A monotone nonlinear load gives one physical operating current + − 14 V R nonlinear I (a) Method and calculation 14.0 = I(1.00) + Vd(I) ⇒I3 + 3I −14 = 0 (I in A). I = 2.00 A, d(I3 + 3I)/dI = 3I2 + 3 > 0. (b) Interpretation and further calculation δE = [1.00 + 2.00 + 3(1.00)I2] δI. I = 2.00 A : δI ≃δE/(15.0 Ω). Physical check A strictly increasing total voltage requirement intersects a fixed source voltage at exactly one current.

Question 5

A monotone nonlinear load gives one physical operating current

An ideal 14.0 V source drives a 1.00 ohm resistor and a nonlinear load with Vd(I)=(2.00Ω)I+(1.00V/A3)I3V_d(I)=(2.00\,\Omega)I+(1.00\,\mathrm{V/A^3})I^3.

Diagram shown in the original SVG worksheet.

  1. Find the steady current and prove uniqueness.

  2. Find the small current change for a small emf change deltaE near that state.

Question 5 - Solution

A monotone nonlinear load gives one physical operating current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

14.0=I(1.00)+Vd(I)⇒I3+3I−14=0(I in A).14.0=I(1.00)+V_d(I)\Rightarrow I^3+3I-14=0\quad(I\text{ in A}).I=2.00A,d(I3+3I)/dI=3I2+3>0.I=2.00\,\mathrm A,\quad d(I^3+3I)/dI=3I^2+3>0.

(b) Interpretation and further calculation

δE=[1.00+2.00+3(1.00)I2]δI.\delta E=[1.00+2.00+3(1.00)I^2]\,\delta I.I=2.00A:δI≃δE/(15.0Ω).I=2.00\,\mathrm A:\quad\delta I\simeq\delta E/(15.0\,\Omega).

Physical check

A strictly increasing total voltage requirement intersects a fixed source voltage at exactly one current.